Question

A bag contains $3$ red and $3$ white balls. Two balls are drawn one by one. The probability that they are of different colours is

• A. $\frac{3}{10}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. None of these

Answer 1

Answer: (C)

$\frac{3}{5}$

Let $A\equiv$ event that drawn ball is red
$B\equiv$ event that drawn ball is white
Then $AB$ and $BA$ are two disjoint cases of the given event
Therefore, $P(AB+BA)=P\left(AB\right)+P\left(BA\right)$
$=P\left(A\right)P\left(\frac{B}{A}\right)+P\left(B\right)P\left(\frac{A}{B}\right)$

$=\frac{3}{6}.\frac{3}{5}+\frac{3}{6}.\frac{3}{5}$

$=\frac{3}{5}$